
# 5.1: Definition and Properties


Definition. Let $$f\in C_0^s(\mathbb{R}^n)$$, $$s=0,1,\ldots$$ . The function $$\hat{f}$$ defined by

\label{four1}
\widehat{f}(\xi)=(2\pi)^{-n/2}\int_{\mathbb{R}^n}\ e^{-i\xi\cdot x}f(x)\ dx,

where $$\xi\in\mathbb{R}^n$$, is called {\it Fourier transform} of $$f$$, and the function $$\widetilde{g}$$
given by

\label{invfour1}
\widetilde{g}(x)=(2\pi)^{-n/2}\int_{\mathbb{R}^n}\ e^{i\xi\cdot x}g(\xi)\ d\xi

is called inverse Fourier transform, provided the integrals on the right hand side
exist.

From (\ref{four1}) it follows by integration by parts that differentiation of a function is changed to multiplication of its Fourier transforms, or an analytical operation is converted into an algebraic operation. More precisely, we have

Proposition 5.1.

$$\widehat{D^\alpha f}(\xi)=i^{|\alpha|}\xi^\alpha\widehat{f}(\xi),$$
where $$|\alpha|\le s$$.

The following proposition shows that the Fourier transform of $$f$$ decreases rapidly for $$|\xi|\to\infty$$, provided $$f\in C_0^s(\mathbb{R}^n)$$. In particular, the right hand side of (\ref{invfour1}) exists for $$g:=\hat{f}$$ if $$f\in C_0^{n+1}(\mathbb{R}^n)$$.

Proposition 5.2. Assume $$g\in C_0^s(\mathbb{R}^n)$$, then there is a constant $$M=M(n,s,g)$$ such that
$$|\widehat{g}(\xi)|\le \frac{M}{(1+|\xi|)^s}.$$

Proof. Let $$\xi=(\xi_1,\ldots,\xi_n)$$ be fixed and let $$j$$ be an index such that
$$|\xi_j|=\max_k |\xi_k|$$. Then
$$|\xi|=\left(\sum_{k=1}^n\xi_k^2\right)^{1/2}\le\sqrt{n}|\xi_j|$$
which implies
\begin{eqnarray*}
(1+|\xi|)^s&=&\sum_{k=0}^s{s\choose k}|\xi|^k\\
&\le&2^s\sum_{k=0}^sn^{k/2}|\xi_j|^k\\
&\le&2^sn^{s/2}\sum_{|\alpha|\le s}|\xi^\alpha|.
\end{eqnarray*}
This inequality and Proposition 5.1 imply
\begin{eqnarray*}
(1+|\xi|)^s|\widehat{g}(\xi)|&\le&2^sn^{s/2}\sum_{|\alpha|\le s}|(i\xi)^\alpha\widehat{g}(\xi)|\\
&\le&2^sn^{s/2}\sum_{|\alpha|\le s}\int_{\mathbb{R}^n}\ |D^\alpha g(x)|\ dx=:M.
\end{eqnarray*}

$$\Box$$

The notation inverse Fourier transform for (\ref{invfour1}) is justified by

Theorem 5.1. $$\widetilde{\widehat{f}}=f$$ and $$\widehat{\widetilde{f}}=f$$.

Proof. See \cite{Yosida}, for example. We will prove the first assertion

\label{four2}
(2\pi)^{-n/2}\int_{\mathbb{R}^n}\ e^{i\xi\cdot x}\widehat{f}(\xi)\ d\xi=f(x)

here. The proof of the other relation is left as an exercise. All integrals appearing in the following exist, see Proposition 5.2 and the special choice of $$g$$.

(i) Formula

\label{four3}
\int_{\mathbb{R}^n}\ g(\xi)\widehat{f}(\xi)e^{ix\cdot\xi}\ d\xi=\int_{\mathbb{R}^n}\ \widehat{g}(y)f(x+y)\ dy

follows by direct calculation:
\begin{eqnarray*}
&&\int_{\mathbb{R}^n}\ g(\xi)\left((2\pi)^{-n/2}\int_{\mathbb{R}^n}\ e^{-ix\cdot y}f(y)\ dy\right)e^{i x\cdot\xi}\ d\xi\\
&&\qquad =(2\pi)^{-n/2}\int_{\mathbb{R}^n}\left(\int_{\mathbb{R}^n}\ g(\xi)e^{-i\xi\cdot(y-x)}\ d\xi\right)f(y)\ dy\\
&&\qquad =\int_{\mathbb{R}^n}\ \widehat{g}(y-x)f(y)\ dy\\
\end{eqnarray*}

(ii) Formula

\label{four4}
(2\pi)^{-n/2}\int_{\mathbb{R}^n}\ e^{-i y\cdot\xi}g(\varepsilon\xi)\ d\xi=\varepsilon^{-n}\widehat{g}(y/\varepsilon)

for each $$\varepsilon>0$$ follows after substitution $$z=\varepsilon\xi$$ in the left hand side of (\ref{four1}).

(iii) Equation

\label{four5}
\int_{\mathbb{R}^n}\ g(\varepsilon\xi)\widehat{f}(\xi)e^{i x\cdot \xi}\ d\xi=\int_{\mathbb{R}^n}\ \widehat{g}(y)f(x+\varepsilon y)\ dy

follows from (\ref{four3}) and (\ref{four4}). Set $$G(\xi):=g(\varepsilon\xi)$$, then (\ref{four3}) implies
$$\int_{\mathbb{R}^n}\ G(\xi)\widehat{f}(\xi)e^{i x\cdot\xi}\ d\xi=\int_{\mathbb{R}^n}\ \widehat{G}(y)f(x+y)\ dy.$$
Since, see (\ref{four4}),
\begin{eqnarray*}
\widehat{G}(y)&=&(2\pi)^{-n/2}\int_{\mathbb{R}^n}\ e^{-iy\cdot\xi}g(\varepsilon\xi)\ d\xi\\
&=&\varepsilon^{-n}\widehat{g}(y/\varepsilon),
\end{eqnarray*}
we arrive at
\begin{eqnarray*}
\int_{\mathbb{R}^n}\ g(\varepsilon\xi)\widehat{f}(\xi)&=&\int_{\mathbb{R}^n}\varepsilon^{-n}\widehat{g}(y/\varepsilon)f(x+y)\ dy\\
&=&\int_{\mathbb{R}^n}\widehat{g}(z)f(x+\varepsilon z)\ dz.
\end{eqnarray*}
Letting $$\varepsilon\to 0$$, we get

\label{four6}
g(0)\int_{\mathbb{R}^n}\ \widehat{f}(\xi)e^{i x\cdot\xi}\ d\xi=f(x)\int_{\mathbb{R}^n}\ \widehat{g}(y)\ dy.

Set
$$g(x):=e^{-|x|^2/2},$$
then

\label{four7}
\int_{\mathbb{R}^n}\ \widehat{g}(y)\ dy=(2\pi)^{n/2}.

Since $$g(0)=1$$, the first assertion of Theorem 5.1 follows from (\ref{four6}) and (\ref{four7}). It remains to show (\ref{four7}).

(iv) Proof of (\ref{four7}). We will show
\begin{eqnarray*}
\widehat{g}(y):&=&(2\pi)^{-n/2}\int_{\mathbb{R}^n}\ e^{-|x|^2/2}e^{-ix\cdot x}\ dx\\
&=&e^{-|y|^2/2}.
\end{eqnarray*}
The proof of
$$\int_{\mathbb{R}^n}\ e^{-|y|^2/2}\ dy=(2\pi)^{n/2}$$
is left as an exercise. Since
$$-\left(\frac{x}{\sqrt{2}}+i\frac{y}{\sqrt{2}}\right)\cdot\left(\frac{x}{\sqrt{2}}+i\frac{y}{\sqrt{2}}\right)=-\left(\frac{|x|^2}{2}+i x\cdot y-\frac{|y|^2}{2}\right)$$
it follows
\begin{eqnarray*}
\int_{\mathbb{R}^n}\ e^{-|x|^2/2}e^{-ix\cdot y}\ dx&=&\int_{\mathbb{R}^n}\ e^{-\eta^2}e^{-|y|^2/2}\ dx\\
&=&e^{-|y|^2/2}\int_{\mathbb{R}^n}\ e^{-\eta^2}\ dx\\
&=&2^{n/2}e^{-|y|^2/2}\int_{\mathbb{R}^n}\ e^{-\eta^2}\ d\eta
\end{eqnarray*}
where
$$\eta:=\frac{x}{\sqrt{2}}+i\frac{y}{\sqrt{2}}.$$
Consider first the one-dimensional case. According to Cauchy's theorem we have
$$\oint_C\ e^{-\eta^2}\ d\eta=0,$$
where the integration is along the curve $$C$$ which is the union of four curves as indicated in Figure \ref{fourfig}.

Figure 5.1.1: Proof of (\ref{four7})

Consequently
$$\int_{C_3}\ e^{-\eta^2}\ d\eta=\frac{1}{\sqrt{2}}\int_{-R}^R\ e^{-x^2/2}\ dx-\int_{C_2}\ e^{-\eta^2}\ d\eta-\int_{C_4}\ e^{-\eta^2}\ d\eta.$$
It follows
$$\lim_{R\to\infty}\int_{C_3}\ e^{-\eta^2}\ d\eta=\sqrt{\pi}$$
since
$$\lim_{R\to\infty}\int_{C_k}\ e^{-\eta^2}\ d\eta=0,\ \ k=2,\ 4.$$
The case $$n>1$$ can be reduced to the one-dimensional case as follows. Set
$$\eta=\frac{x}{\sqrt{2}}+i\frac{y}{\sqrt{2}}=(\eta_1,\ldots,\eta_n),$$
where
$$\eta_l=\frac{x_l}{\sqrt{2}}+i\frac{y_l}{\sqrt{2}}.$$
From $$d\eta=d\eta_1\ldots d\eta_l$$ and
$$e^{-\eta^2}=e^{-\sum_{l=1}^n\eta_l^2}=\prod_{l=1}^ne^{-\eta_l^2}$$
it follows
$$\int_{\mathbb{R}^n}\ e^{-eta^2}\ d\eta=\prod_{l=1}^n\int_{\Gamma_l}\ e^{-\eta_l^2}\ d\eta_l,$$
where for fixed $$y$$
$$\Gamma_l=\{z\in{\mathbb C}:\ z=\frac{x_l}{\sqrt{2}}+i\frac{y_l}{\sqrt{2}}, -\infty<x_l<+\infty\}.$$

$$\Box$$

There is a useful class of functions for which the integrals in the definition of $$\widehat{f}$$ and $$\widetilde{f}$$ exist.

For $$u\in C^\infty(\mathbb{R}^n)$$ we set
$$q_{j,k}(u):=\max_{\alpha:\ |\alpha|\le k}\left(\sup_{\mathbb{R}^n}\left((1+|x|^2)^{j/2}|D^\alpha u(x)|\right)\right).$$

Definition. The Schwartz class of rapidly decreasing functions is
$${\mathcal{S}}(\mathbb{R}^n)=\left\{u\in C^\infty(\mathbb{R}^n): \ q_{j,k}(u)<\infty\ \mbox{for any}\ j,k\in{\mathbb N}\cup\{0\}\right\}.$$

This space is a Frechét space.

Proposition 5.3. Assume $$u\in{\mathcal{S}}(\mathbb{R}^n)$$, then $$\widehat{u}$$ and $$\widetilde{u}\in{\mathcal{S}}(\mathbb{R}^n)$$.

Proof. See [24], Chapter 1.2, for example, or an exercise.

## Contributors

• Integrated by Justin Marshall.